Linear Algebra to solve Autosomal inheritance

Course Title: Linear Algebra and Complex Variables

Course Code: MAT205

Section: 06

Course Instructor: Ahsan Ali

Senior Lecturer (Mathematics)

Presented by: Group [B]

Abdullah al bayezid

Rakib Ahmed

Richita Islam

Md. Shabab Mehebub

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Overview

• Introduction Definitions

Applications of Linear algebra in Genetics

• Applications Autosomal inheritance

Probability of diseases

• Conclusion

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Introduction• Genetics: The study of the inheritance of traits.

• Gene: A section of DNA that influences the heredity of a trait.

• Linear algebra: It is the branch of mathematics concerning vector spaces and linear mappings between such spaces. Transformation of matrix is one of the part of them.

• Autosomal inheritance: Is one pattern of inheritance for a

trait, disease, or disorder to be passed on through families

• Transformation of matrix: In Linear algebra, linear transformation can be represented by matrices.

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Applications of Linear algebra in Genetics

• In humans, color blindness, hereditary baldness, hemophilia are

traits controlled by X-linked inheritance and cystic fibrosis,

sickle-cell anemia and Cooley’s anemia occurred because of

autosomal recessive inheritance.

• Investigation can be occurred for the propagation of an

inherited trait in successive generations by computing powers

of a matrix.

• By calculating eigenvalues, eigenvectors and matrix rotation

autosomal recessive inheritance diseases effectiveness could be

counted.

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Linear Algebra to solve Autosomal inheritance

Suppose that a farmer has a large population of plants consisting of some

distribution of all three possible genotypes AA, Aa, and aa. The farmer desires to

undertake a breeding program in which each plant in the population is always

fertilized with a plant of genotype AA and is then replaced by one of its offspring.

We want to derive an

expression for the distribution of the three possible genotypes in the population

after any number of generations.5

For n = 0, 1, 2, . . . , let us setan = fraction of plants of genotype AA in nth generationbn = fraction of plants of genotype Aa in nth generationcn = fraction of plants of genotype aa in nth generation

Thus a0, b0, and c0 specify the initial distribution of the genotypes. We also have that

an + bn + cn = 1 for n = 0, 1, 2, . . .

From that Table we can determine the genotype distribution of each

generation from the genotype distribution of the preceding generation by

the following equations:

an = an-1 + ½ bn-1

bn = cn-1 + ½ bn-1 n = 1, 2, . . . (1)

cn = 0

Linear Algebra to solve Autosomal inheritance

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Linear Algebra to solve Autosomal inheritance

For example, the first of these three equations states that all the offspring of a

plant of genotype AA will be of genotype AA under this breeding program and

that half of the offspring of a plant of genotype Aa will be of genotype AA.

Equations (1) can be written in matrix notation as

x(n) = Mx(n-1) , n = 1, 2, . . . (2)

where

(3)

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Linear Algebra to solve Autosomal inheritance

Consequently, if we can find an explicit expression for Mn, we can use (3) to obtain

an explicit expression for x(n) . To find an explicit expression for Mn , we first

diagonalize M. That is, we find an invertible matrix P and a diagonal matrix D

such that

M = PDP -1 (4)

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Linear Algebra to solve Autosomal inheritance

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Linear Algebra to solve Autosomal inheritance

(5)

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Linear Algebra to solve Autosomal inheritance

Result: Therefore, an= plants of genotype AA=1 .

So, It will be expressed.

But, as bn =0 and cn=0

So, there genotype respectively Aa and aa will not be expressed.

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Probability of diseases Autosomal inheritance Diseases:

By following the calculation of inheritance in which a normal

gene A dominates an abnormal gene a. Genotype AA is a normal

individual; genotype Aa is a carrier of the disease but is not

afflicted with the disease; and genotype aa is afflicted with the

disease. All the offspring of a plant of genotype AA will be of

genotype AA under this breeding program and that half of the

offspring of a plant of genotype Aa will be of genotype AA.

So, according to the calculation of linear algebra we can see that

there have a few probabilities to occur autosomal recessive

inheritence.

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Conclusion

Possible way to control such a disease is for the breeder to always

mate a female, regardless of her genotype, with a normal male. In

this way, all future offspring will either have a normal father and a

normal mother (AA–AA matings) or a normal father and a carrier

mother (AA–Aa matings).

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Thank You

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